Ex 13.2, 12
A die is tossed thrice. Find the Probability of getting an odd number at least once.
A die is tossed thrice
Probability of getting an odd number at least once
= Probability of getting an odd number once
+ Probability of getting an odd number twice
+ Probability of getting an odd number thrice
= 1 Probability of getting an odd number 0 times
= 1 Probability of getting an even number all 3 times
Let the events be
O : getting an odd number
E : getting an even number
Now, Required Probability = 1 P(EEE)
So, P(EEE) = P(E) P(E|E) P(EE|E)
P(E) = Probability of getting an even number
Even numbers on a die = {2, 4, 6} = 3
Total numbers on a die = {1, 2, , 6} = 6
P(E) = 3 6 = 1 2
P(E|E) = Probability of getting an even number on the second die given even number appeared on the first die
Even numbers on a die = {2, 4, 6} = 3
Total numbers on a die = {1, 2, , 6} = 6
P(E|E) = 3 6 = 1 2
P(EE|E) = Probability of getting an even number on the third die, given even numbers appeared on the first two die.
Even numbers on a die = {2, 4, 6} = 3
Total numbers on a die = {1, 2, , 6} = 6
P(EE|E) = 3 6 = 1 2
So,
P(EEE) = P(E) P(E|E) P(EE|E)
P(EEE) = 1 2 1 2 1 2
= 1 8
Probability of getting an odd number at least once = 1 P(EEE)
= 1 1 8
=

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